The integral _ 0 ^ 1 1 _ 7 ^ [1 x ] d x= where [.] denotes the gr… - Vidyadip

MCQ

The integral $\int_{0}^{1} \frac{1}{{ }_{7}^{\left[\frac{1}{x}\right]}} d x=$ where [.] denotes the greatest integer function is equal to

✓
$1+6 \log _{e}\left(\frac{6}{7}\right)$
B
$1-6 \log _{e}\left(\frac{6}{7}\right)$
C
$\log _{e}\left(\frac{7}{6}\right)$
D
$1-7 \log _{ e }\left(\frac{6}{7}\right)$

✓

Answer

Correct option: A.

$1+6 \log _{e}\left(\frac{6}{7}\right)$

a
$\int\limits_{0}^{1} \frac{1}{7\left[\frac{1}{x}\right]} dx =-\int\limits_{1}^{0} \frac{1}{\left.7^{\left[\frac{1}{x}\right.}\right]} dx$

$=(-1)\left[\int\limits_{1}^{1 / 2} \frac{1}{7} dx +\int\limits_{1 / 2}^{1 / 3} \frac{1}{7^{2}} dx +\int\limits_{1 / 3}^{1 / 4} \frac{1}{7^{3}} dx +\ldots \ldots \infty\right]$

$=\left(\frac{1}{7}+\frac{1}{2 \cdot 7^{2}}+\frac{1}{3 \cdot 7^{3}}+\ldots \infty\right)-\left(\frac{1}{7 \cdot 2}+\frac{1}{7^{2} \cdot 3}+\frac{1}{7^{2} \cdot 4} \ldots \infty\right)$

$=-\ln \left(1-\frac{1}{7}\right)-7\left(\frac{1}{7^{2} \cdot 2}+\frac{1}{7^{3} \cdot 3}+\frac{1}{7^{4} \cdot 4}+\ldots . \ldots\right)$

$= {\left[\ln \frac{6}{7}-7\left(-\ln \left(1-\frac{1}{7}\right)-\frac{1}{7}\right)\right.}$

$=6 \ln \frac{6}{7}+1$

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